By Héron Bernard, Issard-Roch Françoise, Picard Colette

Heron, Issard-Roch, Picard. examine numerique (fr)(Dunod, 1999)(ISBN 2100043722)(C)(302s)

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**Sample text**

That is, there exist f0 ∈ Fn and δ > 0, such that f = f0 + δg with g p = 1. Thus m{T ∗ (f0 + δg) > n} ≤ ε. Then m{T ∗ g > 2n/δ} ≤ m{T ∗ (f0 + δg) > n} + m{T ∗ (f0 − δg) > n} ≤ 2ε. Therefore for every g ∈ Lp [−π, π] m{T ∗ g > (2n/δ) g p } ≤ 2ε. Hence, if we deﬁne C(α) = sup m{T ∗ g > α g p }, the function C(α) satisﬁes limα→+∞ C(α) = 0. This principle is completed with the fact that under the same hypothesis about (Tn ), the set of f ∈ Lp [−π, π], where the limit limn Tn f (x) exists a. , is closed in Lp [−π, π].

N∈Z −π ein(x−t) f (t) dt . x−t This step follows the observation of Luzin. We also want to prove that C ∗ f (x) is in Lp (−π/2, π/2) when f ∈ p L (−π, π). By the interpolation theorems we must bound the measure of the set where C ∗ f (x) > y. v. −π ein(x−t) f (t) dt. x−t The main idea of the proof is to decompose the interval I into subintervals, one of them, I(x), containing x (x ∈ I(x)/2). v. v. x−t I(x) eiλ(x−t) f (t) dt + x−t J J eiλ(x−t) f (t) dt. x−t Clearly, the ﬁrst term gives the main contribution.

6 The conjugate function 25 The operator R is related to the conjugate harmonic function. Consider a power series (an + ibn )z n . n>0 Its real and imaginary part for z = eit are (an cos nt − bn sin nt); u= v= n>0 (an sin nt + bn cos nt). n>0 We say that v = u is the conjugate series to u. The operator H that sends u to v must satisfy H(cos nt) = sin nt; H(sin nt) = − cos nt. It is the same to say H(eint ) = −i sgn(n)eint . The R and H operators are related by R f (θ) + eiθ R e−iθ f (θ) = f (θ) + iH f (θ) .